WAEC SSCE - Physics - 2004 (Essay)
Vraag 1 Verslag
(a) List two properties of cathode rays
(b) Explain how the intensity and energy of cathode rays may be increased
(a) Two properties of cathode rays:
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
Antwoorddetails
(a) Two properties of cathode rays:
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
Vraag 2 Verslag
The diagram above illustrates the path ABC, in a vertical x z plane, of a bullet shot into the air at an angle above the horizontal. Copy the diagram, and, using arrows, indicate the relative magnitudes and directions of the vertical and horizontal components of the velocities of the bullet at the point A, B and C.
What the diagram shows. In the vertical x-z plane the bullet follows the parabolic path from A to B to C. Point A is the launch point where the bullet is rising, B is the highest point of the flight, and C is where the path returns to the horizontal on landing. The x-axis is horizontal and the z-axis is vertical.
Concept being tested. A projectile's velocity is treated as two independent parts: a horizontal component \(v_x\) and a vertical component \(v_z\). The single rule that decides how each behaves is which forces act on the bullet. Ignoring air resistance, the only force in flight is gravity, and gravity acts straight down.
- Because there is no horizontal force, there is no horizontal acceleration, so the horizontal component \(v_x\) stays constant. Its arrow has the same length and the same forward (+x) direction at A, B and C.
- Because gravity acts vertically, it produces a constant downward acceleration \(g\). This makes the vertical component \(v_z\) change continuously: it starts large and upward at A, decreases to zero at the top B, then grows downward on the way to C. By the symmetry of the parabola the upward speed at A equals the downward speed at C.
Required diagram. The path is copied below with the velocity components drawn as arrows. Notice that the three blue horizontal arrows are all the same length (constant \(v_x\)), the red vertical arrow is long and upward at A, absent at B (zero), and equally long but downward at C. The dashed purple arrow is the resultant (actual) velocity, tangent to the path.
Reading the arrows. The relative magnitudes and directions are:
| Point | Horizontal component \(v_x\) | Vertical component \(v_z\) | Resultant velocity |
|---|---|---|---|
| A (rising) | Forward (+x), full length | Upward, maximum: \(v_z = u\sin\theta\) | Points up-and-forward, tangent to the path |
| B (top) | Forward (+x), same length as at A | Zero: \(v_z = 0\) | Purely horizontal (forward) |
| C (falling) | Forward (+x), same length as at A | Downward, maximum, equal in size to A: \(v_z = -u\sin\theta\) | Points down-and-forward, mirror image of A |
So the horizontal arrows satisfy
\[ v_{x,A} = v_{x,B} = v_{x,C} = u\cos\theta \quad (\text{constant, forward}), \]while the vertical component follows
\[ v_{z,A} = +u\sin\theta, \qquad v_{z,B} = 0, \qquad v_{z,C} = -u\sin\theta. \]Common mistake to avoid. Do not shorten the horizontal arrow at the top, and do not make the bullet momentarily "stop" at B. Only the vertical component is zero at the highest point; the bullet is still moving forward at speed \(u\cos\theta\), which is why it keeps travelling and does not fall straight down. Equally, keep the horizontal arrows at A, B and C exactly the same length, because nothing pushes or drags the bullet horizontally once air resistance is ignored.
Examination reminder. Marks here are awarded for showing three equal, forward horizontal arrows and a vertical arrow that is large-up at A, zero at B, and large-down at C, with the A and C vertical arrows drawn the same length to show the up-down symmetry of projectile motion.
Antwoorddetails
What the diagram shows. In the vertical x-z plane the bullet follows the parabolic path from A to B to C. Point A is the launch point where the bullet is rising, B is the highest point of the flight, and C is where the path returns to the horizontal on landing. The x-axis is horizontal and the z-axis is vertical.
Concept being tested. A projectile's velocity is treated as two independent parts: a horizontal component \(v_x\) and a vertical component \(v_z\). The single rule that decides how each behaves is which forces act on the bullet. Ignoring air resistance, the only force in flight is gravity, and gravity acts straight down.
- Because there is no horizontal force, there is no horizontal acceleration, so the horizontal component \(v_x\) stays constant. Its arrow has the same length and the same forward (+x) direction at A, B and C.
- Because gravity acts vertically, it produces a constant downward acceleration \(g\). This makes the vertical component \(v_z\) change continuously: it starts large and upward at A, decreases to zero at the top B, then grows downward on the way to C. By the symmetry of the parabola the upward speed at A equals the downward speed at C.
Required diagram. The path is copied below with the velocity components drawn as arrows. Notice that the three blue horizontal arrows are all the same length (constant \(v_x\)), the red vertical arrow is long and upward at A, absent at B (zero), and equally long but downward at C. The dashed purple arrow is the resultant (actual) velocity, tangent to the path.
Reading the arrows. The relative magnitudes and directions are:
| Point | Horizontal component \(v_x\) | Vertical component \(v_z\) | Resultant velocity |
|---|---|---|---|
| A (rising) | Forward (+x), full length | Upward, maximum: \(v_z = u\sin\theta\) | Points up-and-forward, tangent to the path |
| B (top) | Forward (+x), same length as at A | Zero: \(v_z = 0\) | Purely horizontal (forward) |
| C (falling) | Forward (+x), same length as at A | Downward, maximum, equal in size to A: \(v_z = -u\sin\theta\) | Points down-and-forward, mirror image of A |
So the horizontal arrows satisfy
\[ v_{x,A} = v_{x,B} = v_{x,C} = u\cos\theta \quad (\text{constant, forward}), \]while the vertical component follows
\[ v_{z,A} = +u\sin\theta, \qquad v_{z,B} = 0, \qquad v_{z,C} = -u\sin\theta. \]Common mistake to avoid. Do not shorten the horizontal arrow at the top, and do not make the bullet momentarily "stop" at B. Only the vertical component is zero at the highest point; the bullet is still moving forward at speed \(u\cos\theta\), which is why it keeps travelling and does not fall straight down. Equally, keep the horizontal arrows at A, B and C exactly the same length, because nothing pushes or drags the bullet horizontally once air resistance is ignored.
Examination reminder. Marks here are awarded for showing three equal, forward horizontal arrows and a vertical arrow that is large-up at A, zero at B, and large-down at C, with the A and C vertical arrows drawn the same length to show the up-down symmetry of projectile motion.
Vraag 3 Verslag
State three methods of polarising an unpolarised light
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Antwoorddetails
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Vraag 4 Verslag
(a) Define gravitational field intensity
(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\(^2\) on the horizontal axis;
(i) make a sketch to show the nature of the graph,
(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph
(iii) given that the slope of the graph is 0.25, determine the value for g [Take \(\pi\) = 3.142]
(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\(^{-1}\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the
(i) times of light of the stone
(ii) height of the wall [g = 10ms\(^{-2}\)]
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Antwoorddetails
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Vraag 5 Verslag
Define upper fixed point and lower fixed point as used in thermometry
(b) The electrical resistances of the element in a platinum restistance thermometer at 100\(^o\)C, 0\(^o\) and room temperature are 75.000, 63.000 and 64.992 \(\Omega\) respectively. Use these data to determine the room temperature.
(c) (i) State Boyle's law
(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\(^{-2}\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\(^{-2}\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\(^{-2}\)m, Determine the value of the atmospheric pressure. [g = 10ms\(^{-2}\), density of mecury = 1.36 x 10\(^4\) kg m\(^{-3}\)]
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Antwoorddetails
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Vraag 6 Verslag
(a) Explain;
(i) wave motion
(ii) stationary wave
(b)(i) List four physical properties of a wave
(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave
(iii) List the factors on which the frequency of vibration in a stretched string depends
(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;
(i) frequency
(ii) period
(iii) speed of the wave
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Antwoorddetails
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Vraag 7 Verslag
A parallel beam of unpolarised light is incident on a plane glass surface at an angle of 58\(^{o}\) to the normal. If the reflected beam is completely polarised, calculate the refractive index of the glass.
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
Antwoorddetails
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
Vraag 8 Verslag
(a) Define Young's modulus
(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Antwoorddetails
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Vraag 9 Verslag
(a) Define diffusion
(b) State Graham's law of diffusion
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Antwoorddetails
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Vraag 10 Verslag
(a) State two;
(i) properties of x-rays
(ii) reasons to show that x-rays are waves
(iii) uses of x-rays other than those in medicine;
(iv) hazards of x-rays.
(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\(^4\)V and the current in the tube is 2.00 x 10\(^{-2}\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.
(i) maximum frequency of the emitted radiation
(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck's constant, h = 6.63 x 10\(^{-34}\) Js, electronic charge e = 1.60 x 10\(^{-19}\) C]
(a)(i) Two properties of X-rays: they travel in straight lines at the speed of light and are not deflected by electric or magnetic fields (uncharged); they penetrate matter and affect photographic plates.
(ii) Two reasons showing X-rays are waves: they undergo diffraction (and interference) when passed through crystals; they travel with the speed of light and are part of the electromagnetic spectrum (they are not deflected by fields).
(iii) Two uses of X-rays other than in medicine: detecting flaws/cracks in metal castings and welds (industrial radiography); studying crystal structure by X-ray diffraction (crystallography); security screening of luggage.
(iv) Two hazards of X-rays: they damage/destroy living body cells and tissues; prolonged exposure can cause burns, cancer or genetic mutation.
(b) \(V = 5.00 \times 10^{4}\ \text{V}\), \(I = 2.00 \times 10^{-2}\ \text{A}\).
(i) Maximum frequency: the fastest electrons give up all their energy \(eV\) to a single photon, so \(hf_{max} = eV\):
\[ f_{max} = \frac{eV}{h} = \frac{1.60 \times 10^{-19} \times 5.00 \times 10^{4}}{6.63 \times 10^{-34}} \] \[ f_{max} = \frac{8.0 \times 10^{-15}}{6.63 \times 10^{-34}} = 1.21 \times 10^{19}\ \text{Hz} \](ii) Rate of heat removal: total power supplied is:
\[ P = VI = 5.00 \times 10^{4} \times 2.00 \times 10^{-2} = 1000\ \text{W} \]Only 1% becomes X-radiation, so 99% appears as heat that must be removed:
\[ P_{heat} = 0.99 \times 1000 = 990\ \text{W} \]Heat is removed at the rate of \(990\ \text{J s}^{-1}\) (990 W).
Antwoorddetails
(a)(i) Two properties of X-rays: they travel in straight lines at the speed of light and are not deflected by electric or magnetic fields (uncharged); they penetrate matter and affect photographic plates.
(ii) Two reasons showing X-rays are waves: they undergo diffraction (and interference) when passed through crystals; they travel with the speed of light and are part of the electromagnetic spectrum (they are not deflected by fields).
(iii) Two uses of X-rays other than in medicine: detecting flaws/cracks in metal castings and welds (industrial radiography); studying crystal structure by X-ray diffraction (crystallography); security screening of luggage.
(iv) Two hazards of X-rays: they damage/destroy living body cells and tissues; prolonged exposure can cause burns, cancer or genetic mutation.
(b) \(V = 5.00 \times 10^{4}\ \text{V}\), \(I = 2.00 \times 10^{-2}\ \text{A}\).
(i) Maximum frequency: the fastest electrons give up all their energy \(eV\) to a single photon, so \(hf_{max} = eV\):
\[ f_{max} = \frac{eV}{h} = \frac{1.60 \times 10^{-19} \times 5.00 \times 10^{4}}{6.63 \times 10^{-34}} \] \[ f_{max} = \frac{8.0 \times 10^{-15}}{6.63 \times 10^{-34}} = 1.21 \times 10^{19}\ \text{Hz} \](ii) Rate of heat removal: total power supplied is:
\[ P = VI = 5.00 \times 10^{4} \times 2.00 \times 10^{-2} = 1000\ \text{W} \]Only 1% becomes X-radiation, so 99% appears as heat that must be removed:
\[ P_{heat} = 0.99 \times 1000 = 990\ \text{W} \]Heat is removed at the rate of \(990\ \text{J s}^{-1}\) (990 W).
Vraag 11 Verslag
Give three observations in support of de Broglie's assumption that moving particles behave like waves.
Three observations supporting de Broglie's assumption that moving particles behave like waves:
- Davisson-Germer experiment: electrons scattered from a nickel crystal gave a diffraction pattern with definite intensity peaks, just like waves diffracted by a grating; the wavelength found agreed with \(\lambda = \dfrac{h}{mv}\).
- G. P. Thomson's experiment: electrons passed through a thin metal foil produced concentric diffraction rings on a photographic plate, similar to those given by X-rays, confirming electron diffraction.
- The electron microscope: it works because fast electrons have a very short (wave) wavelength, allowing it to resolve objects far smaller than an optical microscope can.
Antwoorddetails
Three observations supporting de Broglie's assumption that moving particles behave like waves:
- Davisson-Germer experiment: electrons scattered from a nickel crystal gave a diffraction pattern with definite intensity peaks, just like waves diffracted by a grating; the wavelength found agreed with \(\lambda = \dfrac{h}{mv}\).
- G. P. Thomson's experiment: electrons passed through a thin metal foil produced concentric diffraction rings on a photographic plate, similar to those given by X-rays, confirming electron diffraction.
- The electron microscope: it works because fast electrons have a very short (wave) wavelength, allowing it to resolve objects far smaller than an optical microscope can.
Vraag 12 Verslag
(a) (i) Explain electromotive force
(ii) list two sources of electromotive force other than a chemical cell
(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,
(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\(_o\) and V respectively.
(i) Explain the physical meanings of V\(_o\) and V
(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Antwoorddetails
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Vraag 13 Verslag
(a) For a water voltameter, identify the;
(i) electrolyte;
(ii) electrodes;
(iii) substances deposited on the electrodes
(b)State the ratio of the volume of the substances deposited in (a) (iii) above
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Antwoorddetails
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Vraag 14 Verslag
In his first attempt, a long jumper took off from the spring board with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{u^2 sin 2\theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\(^{-2}\)]
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Antwoorddetails
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Vraag 15 Verslag
A Spiral spring of natural length 20.00cm has a scale hanging freely in its lower end. When an object ass 40 g is placed in the pan. its length becomes cm. When the object is replaced with another of 60g, the length becomes 22.05cm. Calculate the mass scale pan. [g = 10 ms\(^{-2}\)]
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
Antwoorddetails
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
